Find the smallest natural $$n$$ such that it’s decimal representation ends at 6 and that if the digit 6 is moved to the beginning of the number (e.i. if the original number was $$x1x2...xm6$$ the new number will be $$6x1x2...xm$$) the resulting number is four times the original.
I've found those numbers by doing the multiplication, I mean, since $$4*x1x2...xm6=6x1x2...xm$$ then $$xm$$ must be 4 and through that way I get the numbers $$153846$$ and $$615384$$. I want to know how to get there more formally.
Let the number formed by the digits after $6$ be $x$ then the number is $10x+6$ and it satisfies \begin{eqnarray*} 4(10x+6)=6 \times 10^k +x \end{eqnarray*} where $k$ is the number of digits in $x$. From the eqqation we have $13x=2(10^k-4)$ so we require $ 10^k \equiv 4 \pmod{13}$. Now calculate the powers of $10$ modulo $13$; $10,9,12,3,\color{red}{4}$. So $k=5$ , $(10^5 -4) /13 = 7692$ and $x=15384$. So $153846 \times 4 = 615384$.